Solving the Basel problem using Gauss's law

  • #1
Hi,
so I came across this video: which shows an interesting way to solve the Basel problem using lighthouses. Imagine a lighthouse that has absolute brightness 1. The apparent brightness then follows an inverse-square law. Now imagine an infinite number line with positive integers only (and 0), with a lighthouse at each integer. The Basel problem then becomes: find the apparent brightness at the 0 point. This is simple enough. Now in the comments, I found this:

You can also use Gauss's law to approach the same solution, rather than a geometric approach. Gauss's law works since a radially symmetric field that's magnitude weakens via the inverse square law has its radius term fall out in a surface integral. This means no matter where the lighthouses are within a sphere of radius R, they can be represented by a single lighthouse of combined magnitude in its center. This also means that same combined lighthouse can be represented by equally spaced, equally lit lighthouses along its boundary. By using this law within a cylinder, and holding the "lighthouse surface density" to be 1/2, you find the surface integral to equal to π^2, and a quarter of the cylinder is π^2/4, the same result as using the geometric method. (The circle is quartered to eliminate lighthouses on the negative side of the number line, and double counting when the number line curves upwards to form the circle)
from Copperbotte.
And I can make neither heads nor tails of it. Anyone has an idea of what this means, or how Gauss's law can solve the Basel problem?

Thanks for answering.
 
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Answers and Replies

  • #2
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Well Gauss's law is used to evaluate the electric field in circumstances where there is a lot of symmetry. The light intensity would be proportional to the square of the electric field. With Gauss's law you integrate the flux over a closed surface, as in ##\int \vec{E}\cdot\vec{n}\:dA##, where ##\vec{E}## is the electric field, ##\vec{n}## is a unit normal to the surface and ##dA## is the surface area element. The integral would be proportional to the charge enclosed within the surface, which in this case would be the number of lighthouses. The method depends on ##\vec{E}## having symmetry, for example on the surface of a sphere or a cylinder. Looking at this problem, I don't see any way to place the lighthouses inside such a surface such that you get the necessary symmetry and still preserve the relationship of ##\vec{E}## to ##\Sigma1/r^2## where ##r## is the distance from the lighthouse.
 
  • #3
So, there'd be absolutely no way to solve this?

I think that the commenter actually tried to curve the number line, so that part of it would form the end caps of a cylinder. The comment is kind of cryptic.
 
  • #5

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